I know it is possibile to define finite sum of ordinals: if $(\alpha_i)_{i\in n}$ is a sequence of lenght $n$, with $n$ in $\omega$, then the symbolism $\sum_{i\in n}\alpha_i$ has a (formal) meaning; however, I did not find any text or article which attributes a (formal) meaning to the symbolism $\sum_{i\in\omega_1}\alpha_i$, where $\alpha_i$ is an ordinal for any $i$ in $\omega_1$: so I elaborated the following lemma, hoping it allows to define a "transfinite" sum.
To follow I will indicate with $\mathbf{Ord}$ and $\mathbf{Lim}_+$ the class of ordinals and the calss of not empty limit ordinals; moreover, if the property $\mathbf P$ is any functional (into the set class) then I will say that the corresponding "function" is a class-function.
Lemma
Let be $\xi$ a (not empty) transfinite sequence of lenght $\lambda$ with range a set of ordinals $X$ and thus let be $\bot$ an operation in $\mathbf{Ord}$: there exists a unique class-function $\mathbf F_\bot$ in $\mathbf{Ord}$ such that the equality
$$
\mathbf F_\bot(\alpha+1)=
\begin{cases}
\mathbf F_\bot(\alpha)\,\bot\,\,\xi(\alpha+1),\text{if $\alpha+1<\lambda$}\\
\mathbf F_\bot(\alpha),\text{ otherwise}
\end{cases}
\tag{1}
\label{eq: successive case}
$$
holds for any $\alpha$ in $\mathbf{Ord}$; moreover the equality
$$
\mathbf F_\bot(\alpha)=\sup_{\beta\in\alpha}\mathbf F_\bot(\beta)
\tag{2}
\label{eq: limit case}
$$
holds whether $\alpha$ is in $\mathbf{Lim}_+$; finally, the equality
$$
\mathbf F_\bot(\alpha)=\xi(\alpha)
\tag{3}
\label{eq: zero case}
$$
holds wheter $\alpha$ is zero.Proof.
Firstly, if $h$ is a function with domain an ordinal $\alpha$ and range an ordinal set $X$ then to follow let's we call it (ordinal) calculus of lenght $\alpha$ or more briefly $\alpha$-calculus. So let be $\mathbf P$ the property which is true when one of the following sentences is it:(a) $h$ is a $0$-calculus and the equality
$$
k=\xi(0)
\tag{4}
\label{eq: P zero}
$$
holds;(b) $h$ is an $(\alpha+1)$-calculus, with $\alpha$ in $\mathbf{Ord}$, such that the equality
$$
k=
\begin{cases}
h(\alpha)\,\bot\,\,\xi(\alpha+1),\text{ if $\alpha+1<\lambda$}\\
h(\alpha),\text{ otherwise}
\end{cases}
\tag{5}
\label{eq: P successive}
$$
holds;(c) $h$ is an $\alpha$ calculus, with $\alpha$ in $\mathbf{Lim}_+$, such that the equality
$$
k=\sup_{\beta\in\alpha}h(\beta)
\tag{6}
\label{eq: P limit}
$$
hods;(d) $h$ is not an ordinal calculus and the equality
$$
k=\emptyset
$$
holds.First of all, let's we observe that $\mathbf P$ is a functional: indeed, any set $h$ is alway an ordinal calculus or otherthing; moreover, is easy to check that any entity into the above equalities is uniquely determinend. So by transfinite recursion theorem there exists a class-function $\mathbf F$ in $\mathbf{Ord}$ such that the equality
$$
\mathbf F=\mathbf G(\mathbf F\restriction_\alpha)
\tag{7}
\label{eq: F value}
$$
holds for any $\alpha$ in $\mathbf{Ord}$, where $\mathbf G$ is the class-function corresponding to $\mathbf P$.Now let's we prove that $\mathbf F\restriction_\alpha$, with $\alpha$ in $\mathbf{Ord}$, is a $\alpha$-calculus.
$———————————–$
First of all, let's we assume $\alpha$ is an ordinal such that $\mathbf F\restriction_\beta$ is a $\beta$-calculus for any $\beta$ in $\alpha$ and thus let's we distinguish the following three cases:
(e) $\alpha$ is empty;
(f) $\alpha$ is successor;
(g) $\alpha$ is limit and not empty;
Case (e): In this case $\mathbf F\restriction_\alpha$ is empty so that it will be trivially a $0$-calculus;
Case (f): Let's we assume $\alpha$ is the successor of $\beta$. Now for any $\theta$ in $\alpha$ the inequality
$$
\theta\le\beta
$$
holds so that $\mathbf F\restriction_\alpha$ will be a $\alpha$-calculus wheter $\mathbf F(\beta)$ is an ordinal: indeed by inductive hypotesis $\mathbf F\restriction_\beta$ is a $\beta$-calculus and so $\mathbf F(\gamma)$ is an ordinal for any $\gamma$ in $\beta$.So we can consider the following cases:
$\beta$ is empty;
$\beta$ is successor;
$\beta$ is limit.
So if $\beta$ is empty then by eq. \eqref{eq: P zero} and eq. \eqref{eq: F value} the equality
$$
\mathbf F(\beta)=\mathbf G(\mathbf F\restriction_\beta)=\mathbf G(\emptyset)=\xi(\emptyset)
$$
holds and thus $\mathbf F(\beta)$ is an ordinal since $\xi(\emptyset)$ is it. After all, if $\beta$ is the successor of any $\gamma$ then $\mathbf F\restriction_\beta$ is a $(\gamma+1)$-calculus and thus by eq. \eqref{eq: P successive} and eq. \eqref{eq: F value} the equality
$$
\mathbf F(\beta)=
\mathbf G(\mathbf F\restriction_\beta)=
\mathbf G(\mathbf F\restriction_{\gamma+1})=
\begin{cases}
\mathbf F(\gamma)\,\bot\,\,\xi(\gamma+1),\text{ if $\gamma+1<\lambda$}\\
\mathbf F(\gamma),\text{ otherwise}
\end{cases}
$$
holds: so $\mathbf F(\beta)$ is an ordinal since $\bot$ is an operation in $\mathbf{Ord}$ or since $\mathbf F(\gamma)$ is an ordinal by hypothesis. Finally if $\beta$ is limit then $\mathbf F$ is an ordinal calculus with lenght a limit ordinal and thus by eq. \eqref{eq: P limit} and \eqref{eq: F value} the equality
$$
\mathbf F(\beta)=\mathbf G(\mathbf F\restriction_\beta)=\sup_{\gamma\in\beta}\mathbf F(\gamma)
$$
holds: so $\mathbf F(\beta)$ is an ordinal since by hypotesis $\mathbf F(\gamma)$ is an ordinal for any $\gamma$ in $\beta$ and since the supremum of orinals is an ordinal too.So we conclude that $\mathbf F(\beta)$ is always an ordinal and thus if by replacement axiom $\mathbf F\restriction_\alpha$ is a function (with domain $\alpha$) then it is an $\alpha$-calculus.
Case (g): In this case $\beta+1$ is in $\alpha$ for any $\beta$ in $\alpha$ so that if by inductive hypothesis $\mathbf F\restriction_{\beta+1}$ is an ordinal calculus then $\mathbf F(\beta)$ would be an ordinal: thus by arbitrariness of $\beta$ we conclude that $\mathbf F\restriction_\alpha$ is a $\alpha$-calculus since by replacement axiom it is a function -with domain $\alpha$.
Finally, we conclude that $\mathbf F\restriction_\alpha$ is always a $\alpha$-calculus so that by transfinite induction we conclude that this is true for any ordinal.
$———————————–$
Now by the same argument advanced in (f) it is not hard to conclude equalities eq. \eqref{eq: successive case}-\eqref{eq: zero case} are true; moreover, by eq. \eqref{eq: successive case}-\eqref{eq: zero case} is not hard to conclude that $\mathbf F$ is unique: finally the statement follows.
Now let be $(\alpha_i)_{i\in 2}$ an ordinal sequence of lenght $2$: by eq. \eqref{eq: successive case} we first observe the equality
$$
\mathbf F_\bot(1)=\mathbf F_\bot(0+1)=\mathbf F_\bot(0)\,\bot\,\,\alpha(0+1)=\alpha_0\,\bot\,\,\alpha_1
$$
holds and thus by eq. \eqref{eq: successive case} we conclude even the equality
$$
\mathbf F_\bot(2)=\mathbf F_\bot(1+1)=\mathbf F_\bot(1)=\alpha_0\,\bot\,\,\alpha_1
\tag{8}
\label{eq: binary summ}
$$
holds, since $2$ is not into the domain $(2)$ of $(\alpha_i)_{i\in 2}$. So by eq. \eqref{eq: binary summ} I put
$$
\sum_{i\in\lambda}\alpha_i:=\mathbf F_+(\lambda)
\tag{9}
\label{def: transfinite sum}
$$
where $(\alpha_i)_{i\in\lambda}$ is a transfinite sequence of ordinals with lenght $\lambda$; finally, we can even put
$$
\prod_{i\in\lambda}\alpha_i=\mathbf F_\cdot(\lambda)
\tag{10}
\label{def: transfinite product}
$$
where $(\alpha_i)_{i\in\lambda}$ is yet a transfinite sequence of ordinals with lenght $\lambda$.
So I would like to know if actaully the above lemma allows to define a "transfinite" sums and products and so if I well proved it: could someone help me, please?
N.B.
Attempting to evaluating the "truth" and so the utility of the lemma I observed what to follow.
Let be $(\alpha_i)_{i\in\omega}$ the $\omega$-calculus with costant value $\omega$: if $\bigwedge$ is the operator correspoding to the ordianl exponentiation $\wedge$ then it seems to me the equality
$$
\bigwedge_{i\in\omega}\alpha_i=\sup_{i\in\omega}\mathbf F_\wedge(i)=\sup\left\{\omega,\omega^\omega,\omega^{\omega^\omega},\dotsc\right\}
$$
holds so that the above lemma allows to give a formal definition of $\epsilon$-number.
Best Answer
(Remark in advance: I’m answering this question after seeing it cross-posted to MathOverflow. I think it’s a good question at the heart and well-suited here, but I guess the reason it didn’t initially get an answer here is because the way you’ve asked it is quite difficult to read — the actual question is rather buried in a long wall of text, which presents its material very verbosely and with non-standard terminology. Writing it more clearly and concisely would have made it a lot more inviting to read and answer.)
Yes, this definition or ordinal sums/products by transfinite recursion makes sense, and agrees with the definition by order-types. I’ve certainly seen it before in folklore; I don’t know a reference off the top of my head and don’t have books to hand, but I’d be surprised if it isn’t in one of the general reference texts on set theory (e.g. Moschovakis, Kunen, Jech). The way I’d see it is via a more general schema of transfinite recursion, which is certainly in those standard texts in the following form or something close to it:
General transfinite recursion schema. Briefly: A function $\newcommand{\restr}{\mathord{\upharpoonright}}\newcommand{\On}{\mathrm{On}} G$ on ordinals can be defined by transfinite recursion, by giving functions $F_\alpha$ that specify $G(\alpha)$ in terms of the earlier values $G \restr_\alpha$. Precisely: Suppose we have a class $X_\alpha$ for each ordinal $\alpha$, and function-classes $F_\alpha : \prod_{\beta < \alpha} X_\beta \to X_\alpha$ for each $\alpha$. Then for any ordinal $\alpha$ there’s a unique function $ g_\alpha \in \prod_{\beta<\alpha} X_\beta$ with $g_\alpha(\beta) = F(g_\alpha\mathord{\upharpoonright}_{\beta})$ for all $\beta$, and putting these together, there’s a unique function-class $G$ with domain $\On$ such that for each $\alpha$, $G(\alpha) = F(G\restr_\alpha) \in X_\alpha$.
Proof. Given $X_\alpha$, $F_\alpha$ for $\alpha \in \On$ as above, say an attempt of length $\alpha$ is a function $g \in \prod_{\beta<\alpha}X_\alpha$ satisfying the given defining condition. Note that for any $\beta < \alpha$, if $g$ is an attempt of length $\alpha$, then $g\restr_\beta$ is an attempt of length $\beta$. Given this, it’s direct by induction on $\alpha$ that for each $\alpha$ there’s a unique attempt $g_\alpha$ of length $\alpha$: any attempt $g_\alpha$ must have $g_\alpha \restr_\beta = g_\beta$ (since by induction $g_\beta$ is the unique attempt at $\beta$), so must have $g_\alpha(\beta) = F(g_\beta)$; so this is the only possibly definition of $g_\alpha$, and we can check that this definition does give an attempt. This completes the proof that there’s a unique attempt for each $\alpha$. Taking the union of these over all $\alpha$ gives the desired function on $\On$. $\Box$
Now ordinal-indexed sum can be defined as you suggest: for each $\alpha$, $\Sigma_\alpha$ is an operation on $\alpha$-indexed sequences of ordinals, i.e. $\Sigma_\alpha : \On^\alpha \to \On$, written as $\sum_{i \in \alpha} \beta_i$, and defined by recursion on $\alpha$ as: $$\begin{aligned} \sum_{i \in \alpha} \beta_i & := \sup_{\alpha' < \alpha} \sum_{i \in \alpha'} \beta_i \qquad \text{($\alpha$ limit)} \\ \sum_{i \in \alpha +1} \beta_i & := (\sum_{i\in\alpha} \beta_i) + \beta_\alpha \end{aligned}$$ This is an instance of the general schema above, with the class $X_\alpha$ taken as $(\On^\alpha \to \On)$: for each $\alpha$, the operator $\Sigma_\alpha$ is defined in terms of the operators $\Sigma_{\alpha'}$ for $\alpha' < \alpha$.
(Note also we didn’t need to split out the case $\alpha=0$ separately: it fits into the general case of limit $\alpha$, with the sum of the empty sequence being $\sup \emptyset = 0$.)
Comparing this with the definition of ordinal-indexed sums by order-types (where $\sum_{i \in \alpha} \beta_i$ is the order-type of the lex ordering on pairs $(i,j)$ where $i \in \alpha$, $j \in \beta_i$), it suffices to show that the order-type definition satisfies the two defining equations of the recursive definition: by uniqueness of recursively-defined functions it then follows that the definitions agree.
The case for tranfinite products is exactly analogous.